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kirza4 [7]
3 years ago
11

A group of friends wants to go to the amusement park. They have no more than $275 to spend on parking and admission. Parking is

$5, and tickets cost $27 per person, including tax. Write and solve an inequality which can be used to determine r, the number of people who can go to the amusement park.
Mathematics
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

275=27x+5

Step-by-step explanation:

10 people can go to the park

27(10)+5=275

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Hector is dividing students into groups for an HR wants to divide the boys and girls so that each group has the same number of t
Roman55 [17]

Answer:

7 groups, 3 boys and 8 girls

Step-by-step explanation:

Let total number of groups be x

and total hikers in one group be y

if number of girls and boys in every group is same, then

\frac{21}{x} + \frac{56}{x} = y

\frac{21 + 56}{x} = y\\x*y = 77 ....(1)

from (1) x will either be 11 or 7

for x = 11 the values won't be real.

For x = 7\\y = \frac{77}{7} \\y = 11

so there will be 7 groups with 11 hikers in each groups and every group will have \frac{21}{7} = 3 boys and \frac{56}{7} = 8 girls

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3 years ago
If f(x) = 2/3 x-6, what is the value of f(-9)?
Keith_Richards [23]

Answer:

-12

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4 0
3 years ago
A team of runners is needed ti run 1/4- mile relay race.if each runner must run 1/16 mile,how many runners will be needed
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Read 2 more answers
A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes
Elena L [17]

Answer:

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

\hat p=\frac{282}{830}=0.340 estimated proportion of people that regularly watch the TV station’s news program

p_o=0.38 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:  

Null hypothesis:p\geq 0.38  

Alternative hypothesis:p < 0.38  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

3 0
3 years ago
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